Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be the irreducible polynomial over $K$ and let $L$ be the splitting field ex of $f$. Why should all the roots be simple or none of them can be simple?

I am looking for a example for an extension $K\le L\le M$ such that $M|K$ is normal doesn't mean that $L|K$ and $M|L$ are normal.

share|improve this question
    
There's a mixup with the order of $K$ and $L$ as well as with $L$ and $M$ in the last sentence. What do you mean by your question? Is your question whether a field extension always has to be separable? –  k.stm Oct 23 '12 at 19:45
    
@K.Stm. : it need not be seperable . –  Theorem Oct 23 '12 at 19:50
    
For your first question (which is entirely separate from the second): if $f$ is irreducible and has a repeated root $\alpha$ in its splitting field, then $\alpha$ is also a root of $f'$, from which it follows that $f' = 0$. But this means $f(x) = g(x^p)$ for some polynomial $g$, so all the roots of $f$ have multiplicity at least $p$ (here $p$ is the characteristic of $K$). –  Justin Campbell Oct 23 '12 at 19:50
    
@JustinCampbell : can you explain me why $f(x)=g(x^p)$, i didn't get the argument . –  Theorem Oct 23 '12 at 19:55
add comment

1 Answer

up vote 1 down vote accepted

Picking up where I left off in the comments: if $f$ is irreducible and has a repeated root in its splitting field, then $f' = 0$. But if we write $f(x) = \sum a_nx^n$, then $0 = f'(x) = \sum na_nx^{n-1}$ means that whenever $a_n \neq 0$ we have $n = 0$ in $K$, which is to say $p | n$, where $p$ is the characteristic of $K$. This says precisely that $f(x) = g(x^p)$ for some polynomial $g$.

For your other question: if $M/K$ is normal then $M/L$ is normal automatically, but $K/L$ need not be. For example, take $M = \mathbb{Q}[2^{1/4},i]$, which is Galois (in particular, normal) with group $D_4$, the dihedral group of the square. The subextension $L = \mathbb{Q}[2^{1/4}]$ is not normal: there is an automorphism of $M$ which sends $2^{1/4} \mapsto 2^{1/4}i \notin L$. Alternatively, one can show that $L$ corresponds via Galois theory to a subgroup in $D_4$ generated by a reflection, which is not normal.

share|improve this answer
    
:You have assumed that the field is of character $p$ what if its character is $0$ ? –  Theorem Oct 23 '12 at 20:31
    
If the characteristic is zero then $f' = 0$ implies $f$ is constant. So irreducible polynomials in characteristic zero never have repeated roots: this is strictly a characteristic $p$ phenomenon. –  Justin Campbell Oct 23 '12 at 23:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.